Edexcel iGCSE Physics -1.28P Force, Momentum, and Time- Study Notes- New Syllabus
Edexcel iGCSE Physics -1.28P Force, Momentum, and Time- Study Notes- New syllabus
Edexcel iGCSE Physics -1.28P Force, Momentum, and Time- Study Notes -Edexcel iGCSE Physics – per latest Syllabus.
Key Concepts:
update
Force, Change in Momentum and Time
When a force acts on an object, it causes a change in momentum. The size of the force depends on how quickly this change in momentum occurs.
This relationship explains the forces involved in collisions, braking, and safety features.
Momentum
Momentum is given by:
\( \mathrm{p = mv} \)
- \( \mathrm{p} \) = momentum (kg m/s)
- \( \mathrm{m} \) = mass (kg)
- \( \mathrm{v} \) = velocity (m/s)
Key Relationship
The relationship between force, change in momentum and time taken is:
\( \mathrm{force = \dfrac{change\ in\ momentum}{time\ taken}} \)
\( \mathrm{F = \dfrac{\Delta p}{\Delta t}} \)
- \( \mathrm{F} \) = force (N)
- \( \mathrm{\Delta p} \) = change in momentum (kg m/s)
- \( \mathrm{\Delta t} \) = time taken (s)
Understanding the Equation
- A larger change in momentum produces a larger force.
- A shorter stopping time produces a larger force.
- Increasing the time taken reduces the force.
This is why safety features increase the time taken during collisions.
Impulse
The term impulse is used to describe change in momentum.
\( \mathrm{impulse = force \times time = change\ in\ momentum} \)
\( \mathrm{Ft = \Delta p} \)
Key Idea
- Forces cause changes in momentum.
- Force depends on both momentum change and time.
- Longer stopping times reduce force.
Important Points to Remember
- Use velocity, not speed.
- Time must be in seconds.
- Include correct units.
Example
A ball of mass \( \mathrm{0.5\ kg} \) is travelling at \( \mathrm{10\ m/s} \). It is brought to rest in \( \mathrm{0.2\ s} \). Calculate the average force acting on the ball.
▶️ Answer / Explanation
Initial momentum:
\( \mathrm{p = mv = 0.5 \times 10 = 5\ kg\,m/s} \)
Final momentum = 0
Change in momentum:
\( \mathrm{\Delta p = 5\ kg\,m/s} \)
Use: \( \mathrm{F = \dfrac{\Delta p}{\Delta t}} \)
\( \mathrm{F = \dfrac{5}{0.2}} \)
\( \mathrm{F = 25\ N} \)
Example
A car of mass \( \mathrm{1200\ kg} \) slows down from \( \mathrm{20\ m/s} \) to \( \mathrm{0\ m/s} \) in \( \mathrm{4\ s} \). Calculate the average braking force.
▶️ Answer / Explanation
Change in momentum:
\( \mathrm{\Delta p = 1200 \times (0 – 20) = -24000\ kg\,m/s} \)
Magnitude of change in momentum = \( \mathrm{24000\ kg\,m/s} \)
Use: \( \mathrm{F = \dfrac{\Delta p}{\Delta t}} \)
\( \mathrm{F = \dfrac{24000}{4}} \)
\( \mathrm{F = 6000\ N} \)